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Irina-Kira [14]
3 years ago
15

What is a "curie"?

Chemistry
1 answer:
V125BC [204]3 years ago
8 0

Answer:

C

Explanation:

The curie specifies the amount of ionizing radiation energy that is emitted from an unstable isotope as it decays.

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Consider the following reaction at constant P. Use the information here to determine the value of ΔSsurr at 355 K. Predict wheth
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Answer:

\Delta S_{surr} = + 0.32113\: kJ/K

Explanation:

Given: Entropy of surrounding: ΔSsurr = ?

Temperature: T= 355 K

The change in enthalpy of reaction: ΔH = -114 kJ

Pressure: P = constant

As we know, ΔH = -114 kJ ⇒ negative

Therefore, the given reaction is an exothermic reaction

Therefore, Entropy of surrounding at <em>constant pressure</em> is given by,

\Delta S_{surr} = \frac{-\Delta H}{T}

\therefore \Delta S_{surr} = -\left (\frac{-114 kJ}{355 K}  \right ) = + 0.32113\: kJ/K > 0

<u><em>In the given reaction:</em></u>

2NO(g) + O₂(g) → 2NO₂(g)

As, the number of moles of gaseous products is less than the number of moles of gaseous reactants.

\therefore \Delta S_{system} <  0

As we know, <em>for a spontaneous process, that the total entropy should be positive.</em>

\Delta S_{total} = \Delta S_{surr} + \Delta S_{system} > 0  

<u>Therefore, at the given temperature,</u>

  • if \Delta S_{surr} > \Delta S_{system} \Rightarrow \Delta S_{total} > 0 then the given reaction is spontaneous
  • if \Delta S_{surr} < \Delta S_{system} \Rightarrow \Delta S_{total} < 0 then the given reaction is non-spontaneous
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